An ice floe with a volume of 5 m3 floats on the surface of the water.

An ice floe with a volume of 5 m3 floats on the surface of the water. Determine the volume of the underwater and water parts.

A buoyant force equal to the weight of the displaced fluid acts on a body immersed in a liquid, and is called the Archimedes force:
Fa = ρ * g * V, where ρ is the density of the liquid, g is the acceleration of gravity of a body raised above the Earth g = 9.8 m / s², V is the volume of the body.
In our case:
Fа = ρl * g * Vpod
The body under the action of gravity presses on the water with the force of gravity:
Fт = m * g
We write down the formula for determining the mass, through the density:
m = ρ * V
we substitute it in the formula for gravity:
Fт = m * g = ρ * V * g
We have a body in a state of equilibrium, which means Fа = Fт:
ρl * g * Vpod = ρ * V * g
Let us express Vsub from this expression:
Vpod = V * ρ / ρzh
We will find the density of ice and water according to the reference book:
ρ = 920 kg / m³
ρzh = 1000 kg / m³
Substitute the numerical values:
Vpod = V * ρ / ρzh = 5 * 920/1000 = 4.6 m³
Let’s find the volume of the above-water part:
Vab = V-Vd = 5-4.6 = 0.4 m³
Answer: the volume of the above-water part is 0.4 m³, the volume of the underwater part is 4.6 m³